JEE MAIN - Mathematics (2021 - 26th August Evening Shift - No. 7)
A fair die is tossed until six is obtained on it. Let x be the number of required tosses, then the conditional probability P(x $$\ge$$ 5 | x > 2) is :
$${{125} \over {216}}$$
$${{11} \over {36}}$$
$${{5} \over {6}}$$
$${{25} \over {36}}$$
Explanation
P(x $$\ge$$ 5 | x > 2) = $${{P(x \ge 5)} \over {P(x > 2)}}$$
= $${{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^5}.{1 \over 6} + ....... + \infty } \over {{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^3}.{1 \over 6} + ...... + \infty }}$$
=$${{{{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 5}} \over {1 - {5 \over 6}}}} \over {{{{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6}} \over {1 - {5 \over 6}}}}} = {\left( {{5 \over 6}} \right)^2} = {{25} \over {36}}$$
= $${{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^5}.{1 \over 6} + ....... + \infty } \over {{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^3}.{1 \over 6} + ...... + \infty }}$$
=$${{{{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 5}} \over {1 - {5 \over 6}}}} \over {{{{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6}} \over {1 - {5 \over 6}}}}} = {\left( {{5 \over 6}} \right)^2} = {{25} \over {36}}$$
Comments (0)
